公路设计中的缓和曲线在CAD中绘制时一直是个难题

1、公路设计中的缓和曲线在CAD中绘制时一直是个难题。本人有一种简单的办法可以做到。望顶。输入导线：pline命令绘制1、2、3各点，其对应坐标如下x1=213.7748 y1=92.1117x2=313.7748 y2=92.1117x3=399.6787 y3=143.3026绘制通过ZH、HZ、QZ、HY和YH各点的与路线导线相切的含缓和曲线的平曲线。 通过计算，五个点的直角坐标为：ZH： X=232.9548 Y=92.1117HY： X=285.3608 Y=94.4667QZ： X=311.8101 Y=99.2371YH： X=336.9780 Y=108.6801HZ： X=383

2、.6319 Y=133.7401使用真样条曲线命令绘制含缓和曲线的平曲线的基本操作步骤如下：Command: Spline（启动真样条曲线命令）object/:232.9548,92.1117（通 过 ZH）Enter point :285.3608,94.4667Close/Fit Tolerance/:311.8101,99.2371Close/Fit Tolerance/:336.9708,108.6801Close/Fit Tolerance/:383.6319,133.7401Close/Fit Tolerance/: Ls R) (setq Ls R)(ll_d);ll_v(def

3、un ll_d()(setq os (getvar osmode)(setvar osmode 0)(setq C (* Ls R)q (- (+ (- (/ Ls 2) (/ (* Ls Ls Ls) 240 R R) (/ (* Ls Ls Ls Ls Ls) 34560 R R R R) (/(* Ls Ls Ls Ls Ls Ls Ls) 8386560 R R R R R R)pt1 (cdr (assoc 10 (entget (car p1)pt2 (cdr (assoc 11 (entget (car p1)pt10(polar pt1 (angle pt1 pt2) (/ (

4、distance pt1 pt2) 2)pt3 (cdr (assoc 10 (entget (car p2)pt4 (cdr (assoc 11 (entget (car p2)pt20(polar pt3 (angle pt3 pt4) (/ (distance pt3 pt4) 2)p (+ (- (/ (* Ls Ls) 24 R) (/ (* Ls Ls Ls Ls) 2688 R R R) (/ (* Ls Ls Ls Ls Ls Ls) 506880 R R R R R)jd (inters pt1 pt2 pt3 pt4 nil)alf1(angle pt10 jd)alf2(

5、angle pt20 jd)alf (- (angle jd pt20) alfl)；setq(if (or ( alf pi) (and ( alf (- 0 pi)(progn(setq id_ -1)(if ( alf pi) (setq alf (- (+ pi pi) alf) (setq alf (abs alf)；progn(progn(setq id_ 1)(if ( (abs(distance jd pt1) (abs(distance jd pt2)(setq ptt1 pt1)(setq ptt1 pt2)；if(setq ptt2 (polar jd alf1 (- 0

6、 x1)(thh p1 ptt1 10)(thh p1 ptt2 11)(if ( (abs(distance jd pt3) (abs(distance jd pt4)(setq ptt3 pt3)(setq ptt3 pt4)；if(setq ptt4 (polar jd alf2 (- 0 x1)(thh p2 ptt3 10)(thh p2 ptt4 11)(if (= id_ 1) (command arc pt5 e pt6 r R) (command arc pt6 e pt5 r R)(setq alfd (angf alf)(setvar osmode os)(command

7、 cmdecho 1)(command text pause pause (strcat 偏角=alfd)(command cmdecho 0)(command text (strcat 半径= (rtos R 2 2)(command text (strcat 切线长= (rtos x1 2 2)(command text (strcat 曲 线长= (rtos Cl 2 2)(command text (command text (strcat 外距=(rtos E 2 2)(strcat 缓和曲线长=(rtos Ls 2 2)；ll_d(defun angf (alf)(setq alf

8、f (angtos alf 1 4)n 1kk (strlen alff)(repeat kk(setq alfn (substr alff n 1)(if (= alfn d)(setq nn n);if(setq n (+ n 1);repeat(strcat (substr alff 1 (- nn 1) % (substr alff nn);angf(defun c:hh(/ p1 p2 pt1 pt2 pt3 pt4 pt5 pt6 pt10 pt20 id_ R V Ls E p3 r1 x y l x0 x1 C jd alf alf1 alf2 q p Cl Ls1 Ls2)(

9、command ucs )(setq p1 nil p2 nil)(while (= p1 nil) (setq p1 (entsel n 拾取第一条直线:)(redraw (car p1) 3)(while (= p2 nil) (setq p2 (entsel n 拾取第二条直线:)(redraw (car p2) 3)(initget 1)(setq R (getdist n 请输入弯道半径 R:)(initget 1 Ls V)(setq p3 (getdist n输入缓和曲线长度(Ls )或设计速度(V)：)(if (= p3 V) (ll_v) (progn (setq ls p3

10、) (ll_d)(princ);eline(defun thh(len pt h)(setq en_data (entget (car len)old_data (assoc h en_data)new_data (cons h pt)en （subst new_data old_data en_data）;setq（entmod en）;thh公路设计中的缓和曲线在CAD中绘制时一直是个难题。 本人有一种简单的办法可以做到。望顶。输入导线：pline命令绘制1、2、3各点，其对应坐标如下x1=213.7748y1=92.1117x2=313.7748y2=92.1117x3=399.6787

11、y3=143.3026绘制通过ZH、HZ、QZ、HY和YH各点的与路线导线相切的含缓和曲线的平曲线。通过计算，五个点的直角坐标为：ZH： X=232.9548 Y=92.1117HY： X=285.3608 Y=94.4667QZ： X=311.8101 Y=99.2371YH： X=336.9780 Y=108.6801HZ： X=383.6319 Y=133.7401使用真样条曲线命令绘制含缓和曲线的平曲线的基本操作步骤如下：Command: Spline （启动真样条曲线命令）object/:232.9548,92.1117通过 ZH）Enter point :285.3608,94.4667Close/Fit Tolerance/:311.8101,99.2371Close/Fit Tolerance/:336.9708,108.6801Close/Fit Tolerance/: